AP EAMCET · Maths · Area Under Curves
The area (in sq units) enclosed by the loop of the curve \(a y^2=x^2(a-x),(a>0)\) is
- A \(2 \pi a^2\)
- B \(\frac{\pi}{3} a^2\)
- C \(\frac{4}{15} a^2\)
- D \(\frac{8}{15} a^2\)
Answer & Solution
Correct Answer
(D) \(\frac{8}{15} a^2\)
Step-by-step Solution
Detailed explanation
Given curve is \(a y^2=x^2(a-x),(a > 0)\) The required area \(\begin{aligned} & =2 \int_0^a x \sqrt{\frac{a-x}{a}} d x \\ & =\frac{2}{\sqrt{a}} \int_0^a x \sqrt{a-x} d x \end{aligned}\) Put \(a-x=t^2\), at \(x=0, t=\sqrt{a}\) and , at \(x=a, t=0\) and \(-d x=2 t d t\) So,…
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